2
$\begingroup$

Possible Duplicate:
The Monty Hall Problem - where does our intuition fail us?

Marliyn vos Savant on September 9, 1990 wrote that the "correct" answer to this question was to switch doors, because switching doors gave you a higher probability of winning the car (2/3 instead of 1/3).

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? — Craig F. Whitaker Columbia, Maryland

The basics of this logic puzzle have been repeated more than once.

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male? — Stephen I. Geller, Pasadena, California

Say that a woman and a man (who are unrelated) each has two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys? My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

The question has even tripped up our very own Jeff Atwood. He posed this question:

Let's say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?

Jeff goes on to argue that it was a simple question, asked in simple language and brushes aside the objections of some that say that the question is incorrectly worded if you want the answer to be 2/3.

Is Marilyn vos Savant right, is the answer to these questions 2/3? (What assumptions do you have to make to come to that answer?)

Is the answer to these questions that you have a 50% chance (of getting the car, the other child being a girl, etc.)? (What assumptions do you have to make to come to that answer?)

Is the correct answer, "I don't know"? (What assumptions have to be clarified to be able to answer the question?)

$\endgroup$

marked as duplicate by user88 Jun 15 '12 at 10:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

migrated from skeptics.stackexchange.com Jun 15 '12 at 10:38

This question came from our site for scientific skepticism.

  • $\begingroup$ This is off topic, perhaps you want to ask on Mathematics... $\endgroup$ – Sklivvz Jun 14 '12 at 6:27
  • 1
    $\begingroup$ Monty Hall is one of those "counterintuitive" things, which is why people have problems understanding it. Here is a video that explains it in a simple way. Plus, the MythBusters have tested it and gave it a Confirmed. $\endgroup$ – Oliver_C Jun 14 '12 at 8:51
  • $\begingroup$ “tripped up our very own Jeff Atwood” – do you have references for that? The link you posted is related, but doesn’t corroborate this particular claim. As for the objections you link to, while cute, they have no basis in statistics. Jeff didn’t invent this problem, he just used it; its solution is well established in statistics. (Not that it matters, as this is entirely off topic.) $\endgroup$ – Konrad Rudolph Jun 14 '12 at 12:08
  • $\begingroup$ @KonradRudolph, it tripped him up because he claimed the answer is 2/3 without realizing that the question as worded is ambiguous, and you cannot come to that conclusion. $\endgroup$ – user1873 Jun 14 '12 at 15:10
  • $\begingroup$ @user1873 The question is not worded ambiguously though. It’s counter-intuitive but unambiguous and completely straightforward once you understand conditional probabilities and the difference between p(A|B) and p(B|A). The blog you link to which claims otherwise is simply wrong. But I’ve in the meantime found a comment by Jeff saying that he initially got the wrong result (i.e. 1/2 instead of 2/3). $\endgroup$ – Konrad Rudolph Jun 14 '12 at 15:52